$T : R^n \rightarrow R^n$ is a linear map 
Given $T : R^n \rightarrow R^n$ is a linear map.

Prove that there exists a real number $\alpha$ that the linear transformation $\alpha I - T$ is isomorphism.
Well, a good start is: I can say that $T$ is a linear map from $R^n \rightarrow R^n$ so the dimensions are equal, but it is still not enough. I need to show something else, like if $\alpha I - T$ is injective.
Ideas?
 A: The eigenvalues of $T$ are invariant under a change of basis, so just choose $\alpha$ such that $\alpha$ is distinct from all the eigenvalues of $T$. Then $\alpha I - T$ is an isomorphism. Alternatively, $T$ is invertible if and only if its characteristic polynomial only has non-zero roots. Check that if $\alpha$ is distinct from the roots of the characteristic polynomial (i.e. distinct from the eigenvalues of $T$), then the characteristic polynomial of $\alpha I - T$ has no zero roots.
A: I will show that there is an ${\alpha}\neq0$ that $I-\alpha T$ is surjective and this is equivalent to main problem.
for ${\alpha}=0$  , $I-\alpha T$ is surjective. now if you change $\alpha$ contiously from zero to positive values then image of every point will move countinously in $\mathbb{R}^n$. 
if you have point like $v$ then 
$$d(v, (I-\alpha T)v)\leq \alpha ||T||\cdot|v|$$
that $||T||$ is norm of $T$ as a linear map. 
now get a cube like $[0,1]^n$ and set $\alpha \leq \frac{1}{1000||T||}$. if for this $\alpha$ , $I-\alpha T$ not be surjective then its image should be of dimention less than or equal $n-1$. now of you find distance between vertices of cube and their image it should be a vertex that it's image should be greater than or equal $\frac{1}{2}$ but by the follow inequality this distance should be less than or equal $\frac{1}{1000}$.
A: Since the domain is the same as the codomain, saying that a linear map is not an isomorphism is equivalent to saying it's not injective. 
We'll show by induction that, for all $k\ge1$, there exists a linearly independent set $\{v_1,\dots,v_k\}$ in $\mathbb{R}^n$, where $T(v_i)=\alpha_iv_i$, and the numbers $\alpha_i$ $(i=1,2,\dots,k)$ are pairwise distinct.
This is a contradiction, because a linearly independent set can contain at most $n$ elements.
By assumption, $0I-T$ is not injective, so there is $v_1\ne0$ such that $T(v_1)=0$. The set $\{v_1\}$ is linearly independent and $T(v_i)=0v_i$.
Suppose now we are given a linearly independent set $\{v_1,\dots,v_k\}$ with $T(v_i)=\alpha_iv_i$ and the numbers $\alpha_i$ $(i=1,2,\dots,k)$ are pairwise distinct. Choose $\alpha_{k+1}\ne\alpha_i$ $(i=1,2,\dots,k)$, which is possible because $\mathbb{R}$ is infinite. Since, by assumption, $\alpha_{k+1}I-T$ is not injective, there is $v_{k+1}\ne0$ such that $\alpha_{k+1}v_{k+1}=T(v_{k+1})$. I claim that $\{v_1,\dots,v_k,v_{k+1}\}$ is linearly independent.
Suppose $c_1v_1+\dots+c_kv_k+c_{k+1}v_{k+1}=0$. We have
$$
0=T(0)=T(c_1v_1+\dots+c_kv_k+c_{k+1}v_{k+1})=
c_1\alpha_1v_1+\dots+c_k\alpha_kv_k+c_{k+1}\alpha_{k+1}v_{k+1}
$$
but also
$$
0=\alpha_{k+1}(c_1v_1+\dots+c_kv_k+c_{k+1}v_{k+1})
$$
Subtract the two relations to get
$$
0=c_1(\alpha_1-\alpha_{k+1})v_1+\dots+c_k(\alpha_k-\alpha_{k+1})v_k
$$
that, by the induction hypothesis, implies
\begin{align}
&c_1(\alpha_1-\alpha_{k+1})=0\\
&c_2(\alpha_2-\alpha_{k+1})=0\\
&\vdots\\
&c_k(\alpha_k-\alpha_{k+1})=0
\end{align}
and, since $\alpha_i-\alpha_{k+1}\ne0$ $(i=1,\dots,k)$,
$$
c_1=c_2=\dots=c_k=0
$$
which in turn gives $c_{k+1}v_{k+1}=0$ and so $c_{k+1}=0$. Therefore $\{v_1,\dots,v_k,v_{k+1}\}$ is linearly independent, as claimed.
Note This is modelled on the proof that eigenvectors relative to distinct eigenvalues are linearly independent. However it doesn't require the theory of eigenvalues and eigenvectors.
